Question

Solve: $\int{\dfrac{x}{{{x}^{4}}-1}dx}$ .

Answer 1

Hint: We can see in our problem that, if we use a suitable substitute for our variable ‘x’ in the denominator, the problem could be easily solved. Thus, we will use the substitution method of solving an integral to get the answer to our question. This will make our problem easier and less calculative.

Complete step by step solution:
Let us denote the expression given to us with ‘y’, such that we need to find the integral of ‘y’ over a very small change in ‘x’. Mathematically, this could be written as:
$\begin{align}
  & \Rightarrow I=\int{y.dx} \\
 & or,I=\int{\dfrac{x}{{{x}^{4}}-1}dx} \\
\end{align}$
Where, “I” is the result of this integral.
Let us now substitute the variable ‘${{x}^{2}}$ ’ with the term ‘u’. Thus, the differential of ‘x’ shall also change. This can be calculated as follows:
$\begin{align}
  & \Rightarrow d\left( {{x}^{2}} \right)=d\left( u \right) \\
 & \Rightarrow 2xdx=du \\
 & \therefore xdx=\dfrac{du}{2} \\
\end{align}$
Thus, putting the new substituted values of ‘x’ in terms of ‘u’, our new equation becomes:
$\Rightarrow I=\int{\dfrac{du}{2\left( {{u}^{2}}-1 \right)}}$
The numerator in our integral can be written as:
$\Rightarrow 1.du=\dfrac{\left( u+1 \right)-\left( u-1 \right)}{2}$
And the denominator in our integral could be written as:
$\Rightarrow 2\left( {{u}^{2}}-1 \right)=2\left( u+1 \right)\left( u-1 \right)$
Therefore, our integral now becomes:
$\begin{align}
  & \Rightarrow I=\int{\dfrac{\left( u+1 \right)-\left( u-1 \right)}{4\left( u+1 \right)\left( u-1 \right)}du} \\
 & \Rightarrow I=\dfrac{1}{4}\left[ \int{\dfrac{\left( u+1 \right)}{\left( u+1 \right)\left( u-1 \right)}du}-\int{\dfrac{\left( u-1 \right)}{\left( u+1 \right)\left( u-1 \right)}du} \right] \\
 & \Rightarrow I=\dfrac{1}{4}\left[ \int{\dfrac{1}{\left( u-1 \right)}du}-\int{\dfrac{1}{\left( u+1 \right)}du} \right] \\
\end{align}$
Using the formula for integral of a reciprocal linear function, that is equal to:
$\Rightarrow \int{\dfrac{d\theta }{\theta }}=\ln \theta +C$
Our, equation gets simplified to:
$\begin{align}
  & \Rightarrow I=\dfrac{1}{4}\left[ \int{\dfrac{1}{\left( u-1 \right)}du}-\int{\dfrac{1}{\left( u+1 \right)}du} \right] \\
 & \Rightarrow I=\dfrac{1}{4}\left[ \ln \left( u-1 \right)-\ln \left( u+1 \right) \right]+C \\
 & \Rightarrow I=\dfrac{1}{4}\ln \left( \dfrac{u-1}{u+1} \right)+C \\
\end{align}$
Resubstituting the introduced variable ‘u’ in terms of original variable ‘x’, we get the final result as:
$\Rightarrow I=\dfrac{1}{4}\ln \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+C$
Hence, the integral of $\left[ \int{\dfrac{x}{{{x}^{4}}-1}dx} \right]$ comes out to be $\left[ \dfrac{1}{4}\ln \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+C \right]$ .

Note: Whenever using the substitution method to find out the integral of an expression, we should always be careful as to what to substitute in the place of our original variable. The substitute should be such that it reduces the complexity in our equation. This makes our new equation easy to understand and solve.